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# If root{3-2x} = root(2x) +1, then 4x^2 =

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If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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05 Apr 2011, 20:34
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Difficulty:

75% (hard)

Question Stats:

59% (01:14) correct 41% (02:16) wrong based on 95 sessions

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If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-root-3-2x-root-2x-1-then-4x-135539.html

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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05 Apr 2011, 21:26
=>(3-2x)^1/2=(2x)^1/2+1

Square both sides

=>(3-2x)=(2x)+1+2*(2x)^1/2

=>(2-4x)=2*(2x)^1/2

=>(1-2x)=(2x)^1/2

Square both sides again

1 - 4x + 4x^2 = 2x
4x^2=6x-1
So E.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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05 Apr 2011, 21:28
GMATD11 wrote:
16.) \sqrt{3-2x}=\sqrt{2x}+1 then 4x^2 =

a) 1
b) 4
c) 2-2x
d) 4x-2
e) 6x-1

Guys i solved and got 1

but OA is different .

I want to know the reason nd different approach that i need to follow.

Alternately, to verify your ans 4x^2 = 1 means x = 1/2
Substitute the value 1/2 in eq, it doesn't make LHS=RHS
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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05 Apr 2011, 22:34
root(3-2x) - root(2x) = 1

Squaring both sides :

3 - 2x + 2x - 2 * root(6x - 4x^2) = 1

=> -2 * root(6x - 4x^2) = -2

=> root(6x - 4x^2) = 1

Squaring again :

=> 6x - 4x^2 = 1

=> 4x^2 = 6x - 1

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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06 Apr 2011, 06:17
Squaring on both sides we have
2-4x = 2*sqrt(2x)
Squaring on both sides again we have
4x^2 = 6x-1

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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21 Jun 2014, 01:33
Before proceeding with the solving part , give a look at the options provided. Since none of the options has under root involved in the answer choices think to eliminate that from the target answer.

If u keep the squareroot value on one side w/o making any equation will help us to solve.
squaring both sides and bringing the non squareroot values on the one side and keeping squareroot 2 on the other side .. squaring the terms on both sides ... will give answer ....

let me know in case of any query ...

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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21 Jun 2014, 03:59
GMATD11 wrote:
If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

$$\sqrt{3-2x} = \sqrt{2x} +1$$ --> square both sides: $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ --> $$3-2x=2x+2*\sqrt{2x}+1$$ --> rearrange so that to have root at one side: $$2-4x=2*\sqrt{2x}$$ --> reduce by 2: $$1-2x=\sqrt{2x}$$ --> square again: $$(1-2x)^2=(\sqrt{2x})^2$$ --> $$1-4x+4x^2=2x$$ --> rearrange again: $$4x^2=6x-1$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-root-3-2x-root-2x-1-then-4x-135539.html
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 21 Jun 2014, 03:59
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